Download Reteach 8.3

Name ________________________________________ Date __________________ Class__________________
LESSON
8-3
Reteach
Solving Right Triangles
Use the trigonometric ratio sin A = 0.8 to determine which angle of the triangle
is ∠A.
sin ∠1 =
=
leg opposite ∠1
hypotenuse
sin ∠2 =
6
10
=
= 0.6
leg opposite ∠2
hypotenuse
8
10
= 0.8
Since sin A = sin ∠2, ∠2 is ∠A.
If you know the sine, cosine, or tangent of an acute angle measure, then you can
use your calculator to find the measure of the angle.
Inverse Trigonometric Functions
Symbols
Examples
sin 30° =
sin A = x ⇒ sin−1 x = m∠A
cos 45° =
cos B = x ⇒ cos−1 x = m∠B
tan C = x ⇒ tan−1 x = m∠C
1
⎛ 1⎞
⇒ sin−1 ⎜ ⎟ = 30°
2
⎝2⎠
⎛ 2⎞
2
⇒ cos−1 ⎜
= 45°
⎜ 2 ⎟⎟
2
⎝
⎠
tan 76° ≈ 4.01 ⇒ tan−1 (4.01) ≈ 76°
Use the given trigonometric ratio to determine which angle of the triangle
is ∠A.
1. sin A =
1
2
2. cos A =
_________________________________________
3. cos A = 0.5
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4. tan A =
_________________________________________
13
15
15
26
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Use your calculator to find each angle measure to the nearest degree.
5. sin−1 (0.8)
6. cos−1 (0.19)
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7. tan−1 (3.4)
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________________________________________
⎛ 1⎞
8. sin−1 ⎜ ⎟
⎝5⎠
________________________________________
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8-22
Holt McDougal Geometry
Name ________________________________________ Date __________________ Class__________________
LESSON
8-3
Reteach
Solving Right Triangles continued
To solve a triangle means to find the measures of all the angles and all the sides
of the triangle.
Find the unknown measures of UJKL.
Step 1: Find the missing side lengths.
sin 38° =
← leg opposite ∠K
← hypotenuse
JL
22
13.54 mm ≈ JL
JL2 + LK 2 = JK 2
2
13.542 + LK = 22
2
LK ≈ 17.34 mm
Pythagorean Theorem
Substitute the known values.
Simplify.
Step 2: Find the missing angle measures.
m∠J = 90° − 38°
= 52°
Acute ∠s of a rt. U are complementary.
Simplify.
So JL ≈ 13.54 mm, LK ≈ 17.34 mm, and m∠J = 52°.
Find the unknown measures. Round lengths to the nearest hundredth
and angle measures to the nearest degree.
9.
10.
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________________________________________
_________________________________________
________________________________________
11.
12.
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________________________________________
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________________________________________
For each triangle, find the side lengths to the nearest hundredth and
the angle measures to the nearest degree.
14. J(2, 3), K(−1, −4), L(−1, 3)
13. M(−5, 1), N(1, 1), P(−5, 7)
_________________________________________
________________________________________
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
8-23
Holt McDougal Geometry
Measure
of ∠T
Length
US
Possible
answer:
63°
Possible Possible Possible
answer: answer: answer:
1.96
3
15
1
1
4
16
inches
inches
Tan T
Length
TS
17. 13.99 ft
Sec T
11.33 ft
Possible
answer:
2.20
39°
18. 8 km
62°
28°
6. Possible answer:
19.
The length of US is close in value to tan
T, and the length of TS is close in value to
sec T.
Problem Solving
1. 4.80 ft
2. 9.49 cm
3. 61.4 cm
5. C
7. A
4. 19 in2
6. G
8. F
20. ∠Y
Practice B
Reading Strategies
1. Answers will vary.
2. Both ratios have Hypotenuse in the
denominator.
3. hypotenuse
10
; 0.38
4. a.
26
24
b.
; 0.92
26
10
; 0.42
c.
24
2. ∠1
3. ∠2
4. ∠2
5. ∠2
6. ∠1
7. 55°
8. 24°
9. 79°
10. 22°
11. 77°
12. 6°
14. EF = 2.73 m; m∠D = 65°; m∠F = 25°
15. GH = 7.64 ft; GI = 7.91; m∠I = 44°
16. KL = 2.71 yd; JK = 2.84 yd; m∠K = 17°
17. QP = 11.18 cm; m∠Q = 42°; m∠R = 48°
18. ST = 3.58 yd; m∠S = 12°; m∠T = 78°
Practice A
−1
1. ∠1
13. AB = 7.74 in.; m∠A = 57°; m∠B = 33°
8-3 SOLVING RIGHT TRIANGLES
1. m∠A
21. 3; 5; 5.83
19. BC = 8.60; BD = 7; CD = 5; m∠B = 36°;
m∠C = 54°; m∠D = 90°
2. x
3. tan x
4. ∠2
5. ∠1
6. ∠1
7. ∠1
8. ∠2
9. ∠2
10. 19°
11. 62°
12. 50°
13. 64°
14. 78°
20. LM = 2; LN = 7; MN = 7.28; m∠L = 90°;
m∠M = 74°; m∠N = 16°
21. XY = 1; XZ = 1.41; YZ = 1; m∠X = 45°;
m∠Y = 90°; m∠Z = 45°
Practice C
1. 37°; 53°
2. 23°; 67°
15. 70°
3. 28°; 62°
4. 16°; 74°
16. 3 yd
5. Yes, a 20% grade in the United States is
equal to a 20% grade elsewhere.
Possible answer: A 20% grade in the
37°
53°
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A13
Holt McDougal Geometry
United States means a rise in elevation of
20 feet over 100 horizontal feet. A 20%
grade elsewhere means a rise in
elevation of 20 meters over 100 meters.
20 ft
20 m
But a grade is a ratio:
=
.
100 ft 100 m
The units cancel out, and either way a
1
20% grade simplifies to , or an angle
5
with the horizontal that measures about
11°.
6. EG = 0.61 m; FG = 0.54 m; m∠E = 57°
7. KM = 56.13 mm; m∠K = 61°; m∠L = 34°
6. cos R =
2
; tan R = 1; m∠R = 45°
2
7. sin K =
3
1
; cos K = ; m∠K = 60°
2
2
Problem Solving
1. 16°
2. 22° to 27°
3. 64°
4. 34.9 ft
5. A
6. G
7. D
8. F
Reading Strategies
10. IJ = 5.32 yd; m∠H = 90°; m∠I = 62°
1. finding the measures of all unknown
sides and angles of the triangle
⎛9⎞
2. m∠B = tan−1 ⎜⎝ 4 ⎟⎠
11. RS = 18.98 mm; ST = 20.07 mm; m∠R =
60°
3. sin 24° =
8. BC = 3.74 ft; m∠B = 83°; m∠D = 48°
9. TV = 8.43 in.; UV = 14.08 in.; m∠T = 79°
Reteach
1. ∠2
2. ∠2
3. ∠1
4. ∠2
5. 53°
6. 79°
7. 74°
8. 12°
Practice A
1. horizontal; above
2. depression; below
3. angle of depression
11. QR ≈ 20.76 km; QS ≈ 25.04 km; m∠Q =
34°
4. angle of elevation
5. angle of depression
12. WX ≈ 18.30 cm; m∠X ≈ 59°; m∠Y ≈ 31°
6. angle of elevation
13. MP = MN = 6; PN ≈ 8.49; m∠M = 90°; m∠P =
45°; m∠N = 45°
7. 28 feet
9. 11.9 meters
8. 35 feet
10. 1.9 meters
Practice B
1. angle of elevation
Challenge
3
5
4. 67°; 23°; 12
DEPRESSION
10. FH ≈ 9.12 mi; m∠F ≈ 26°; m∠H ≈ 64°
1.
AB
8-4 ANGLES OF ELEVATION AND
9. AB ≈ 12.52 ft; AC ≈ 7.54 ft; m∠B = 37°
14. KL = 7; LJ = 3; JK ≈ 7.62; m∠L = 90°; m∠J ≈
67°; m∠K ≈ 23°
4
2.
2. angle of depression
4
3
3. angle of depression
3. 53.1°
5
2 5
4. sin E =
; cos E =
; m∠E ≈ 26.6°
5
5
1
3
5. sin M = ; tan M =
; m∠M = 30°
2
3
4. angle of elevation
5. 34 ft 2 in.
6. 37 ft 1 in.
7. 31 ft 10 in.
8. 1.8 m
9. 65°
10. Mr. Shea lives above Lindsey.
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A14
Holt McDougal Geometry